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photoshop1234 [79]
3 years ago
9

The speed of sound at sea level is normally about 340 m/s. A stationary fire alarm has a frequency of 15,000 Hz. An observer run

ning towards the fire alarm hears a frequency of 15,300 Hz. What is the velocity of the observer?
Physics
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

The velocity of the observer is 6.8 m/s

Explanation:

Doppler effect equation is given by the formula:

f'=\frac{v+v_o}{v-v_s} f\\\\where\ f'\ is \ the\ observed \ frequency=15300\ Hz,v_o\ is\ the\ velocity\ of\ the\ observer,\\v_s\ is\ the\ velocity\ of \ source=0,v\ is\ the\ velocity\ of\ sound=340\ m/s\ and\ f\ is\ actual\\frequency=15000\ Hz.\\\\substituting:\\\\15300=15000(\frac{340+v_o}{340} )\\\\340+v_o=346.8\\\\v_o=6.8\ m/sThe velocity of the observer is 6.8 m/s

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Is the property of matter in which a substance can transfer heat or electricity
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4 years ago
A centrifugal pump discharges 300 gpm against a head of 55 ft when the speed is 1500 rpm. The diameter of the impeller is 15.5 i
Ksivusya [100]

Answer: Question 1: Efficiency is 0.6944

Question 2: speed of similar pump is 2067rpm

Explanation:

Question 1:

Flow rate of pump 1 (Q1) = 300gpm

Flow rate of pump 2 (Q2) = 400gpm

Head of pump (H)= 55ft

Speed of pump1 (v1)= 1500rpm

Speed of pump2(v2) = ?

Diameter of impeller in pump 1= 15.5in = 0.3937m

Diameter of impeller in pump 2= 15in = 0.381

B.H.P= 6.0

Assuming cold water, S.G = 1.0

eff= (H x Q x S.G)/ 3960 x B.H.P

= (55x 300x 1)/3960x 6

= 0.6944

Question 2:

Q = A x V. (1)

A1 x v1 = A2 x V2. (2)

Since A1 = A2 = A ( since they are geometrically similar

A = Q1/V1 = Q2/V2. (3)

V1(m/s) = r x 2π x N(rpm)/60

= (0.3937x 2 x π x 1500)/2x 60

= 30.925m/s

Using equation (3)

V2 = (400 x 30.925)/300

= 41.2335m/s

To rpm:

N(rpm) = (60 x V(m/s))/2 x π x r

= (60 x 41.2335)/ 2× π × 0.1905

= 2067rpm.

6 0
3 years ago
To get an idea of the size of magnetic fields at the atomic level, consider the magnitude of the magnetic field due to the elect
artcher [175]

Answer:

The magnetic field is 14.08 T.

Explanation:

Given that,

Speed of electron v=2.2\times10^{6}\ m/s

Distance d =0.5\times10^{-10}\ m

Suppose we need to find the magnetic field at  the location of the proton

We need to calculate the magnetic field

Using formula of magnetic field

B = \dfrac{\mu_{0}I}{2r}....(I)

Using formula of current

I=\dfrac{q}{T}

Using formula of time

T= \dfrac{2\pi r}{v}

Put the value of current and time in equation (I)

B=\dfrac{\mu \times q\times v}{2\pi r\times2r}

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times2.2\times10^{6}}{4\pi\times(0.5\times10^{-10})^2}

B=14.08\ T

Hence, The magnetic field is 14.08 T.

3 0
3 years ago
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